How do you find the roots, real and imaginary, of $y=(x-5)(x-2)$ using the quadratic formula?

Question

How do you find the roots, real and imaginary, of $y=(x-5)(x-2)$ using the quadratic formula?

1 Answer

Impact of this question

1669 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-25T14:42:23

$x_1=5+0i$

$x_2=2+0i$

Explanation:

$y=(x-5)(x-2)=x^2-7x+10$

Let $y=0$

$x^2-7x+10=0$

Using Descartes' rule for $f(x)$ , we know that there are two positive roots. Using it for $f(-x)$ , we know that there can be zero or two negative roots.

$Delta=(-7)^2-4(10)(1)=49-40=9$

$x=(7+-sqrt(9))/2=(7+-3)/2=5, 2$

There are no negative roots and a quadratic can only have, at most, two real roots.

So the real values of $x$ are $5$ and $2$ . These are also the complex roots are we can write them as $5+0i$ and $2+0i$ .

Science

Math

Humanities

... and beyond

How do you find the roots, real and imaginary, of $y=(x-5)(x-2)$ using the quadratic formula?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the roots, real and imaginary, of y=(x-5)(x-2)y=(x-5)(x-2) using the quadratic formula?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the roots, real and imaginary, of $y=(x-5)(x-2)$ using the quadratic formula?